y''-y' = e^3x general solution
You asked:
Investigate the equation: \(y^{\prime \prime} - y^{\prime} = {e}^{3} x\).
MathBot Answer:
\[y{\left(x \right)} = C_{2} e^{x} - x e^{3} - \frac{x^{2} e^{3}}{2} + C_{1}\]
\(C_{1}\), \(C_{2}\) are arbitrary constants.
\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).